Question

Using the following information determine the boiling point of a mixture contains 1560 gm benzene and 1125gm chlorobenzene, when the external pressure is 1000torr Assume the solution is ideal.Given : Molar mass of benzene =78 ; Molar mass of chlorobenzene = 112.5

Answer 1

Answer:

100 C

Explanation:

• $X_{benzene}=\frac{1560/78}{1560/78+1125/112.5} = \frac{2}{3}$
• $X_{chlorobenzene}=\frac{1}{3}$
• At boiling point, the solution's vapour pressure should be 1000 torr.
• According to the graph, at 100 degrees Celsius,
• $P_{chlorobenzene}=250*\frac{1}{3} torr$
• Also, $P_{benzene}=\frac{2}{3}*1300$
• At 100 degrees Celsius, the total vapour of the solution is
• $\frac{250}{3} + \frac{2600}{3}=950 torr$
• The closest pressure to 950 torr is 1000 torr.
• As a result, the right answer is 100 degrees.

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Answer 2

Answer:

100°C

Explanation:

As per the data given,

Mass Benzene= 1560 g

Mass Chlorobenzene = 1125 g

Molar mass Benzene= 78

Molar mass Chlorobenzene = 112.5

External pressure = 1000 torr

Number of moles of Benzene = 1560/78= 20

Number of moles of Chlorobenzene= 1125/112.5= 10

Mole fraction of Benzene(x₁)=20/30=2/3

Mole fraction of Chlorobenzene(x₂) =10/30=1/3

The boiling point of the mixture will be the temperature at which

Total pressure = External pressure

By Raoult’s law,

P= P₁x₁ + P₂x₂

Let the boiling point be 90⁰C :

P= 1000*2/3 + 200*1/3

P = 2200/3

So, here

Total pressure < External pressure

This temperature is not boiling point.

Let the boiling point be 100⁰C :

P = 1350*2/3 + 300*1/3

P = 1000 torr

So, this temperature is the boiling point of the mixture ;

as, Total pressure = External pressure = 1000 torr

The answer is 100⁰C.

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